Difference between revisions of "HRRL Positron Rotating W Target"

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: <math>200 \frac{\mbox{mA}}{\mbox{pulse}}</math>
: <math>200 \frac{\mbox{mA}}{\mbox{pulse}} \times 300 \frac{\mbox{pulses}}{\mbox{sec} \times 0.1 \mu \mbox{s}</math>
Time on for one second is 30 us. and Time off is 0.999999 s The average current 0.999999 * 0 + 0.000001*300*200mA =  
Time on for one second is 30 us. and Time off is 0.999999 s The average current 0.999999 * 0 + 0.000001*300*200mA =  

Revision as of 04:36, 18 March 2011

Q: Does the W target has to be in the vacuum? 
   What if we have it in the air, at the ends of the beam line we have very thin vacuum window? 
   In photofission experiment we had 1 mil vacuum window. under 
   15 MeV e- beam, only 1/1000 electron will interact with the window, produce bremsstrahlung photon.
Q: Can we make positron target that is inner part of the vacuum box, then we rotate all the box. Heat will be transfered to box, 
   box can cooled by water from outside. 
   May be box does not have to rotation movement? I can do other moments like up-down and right-left.

  I think we can make a Tungsten target as a part of the beam line. That means outer part Tungsten target will be beam pine. 
Then we can use water cooling to cool our Tungsten target from outside. 
Before we had all the electrons hitting Faraday Cup, which did not melt the copper even when there was not cooling water 
circling around FC. 
Tungsten has higher melting point, plus we run cooling water around Tungsten target, then Tungsten shouldn't melt.

The first version of a positron converter target will be designed to distribute the heat load by rotating the tungsten target.

Calculate for 1 mm and 2 mm thick Tungsten

Look for Tungsten disks to attach to brushless motor and fit into beam pipe

Beam Heat Load

Accelerator Settings
beam energy 10 MeV
Rep Rate: 300 Hz
I_peak: 200 mA/pulse
pulse width: 0.1 [math]\mu[/math]s (100 ns)
[math]200 \frac{\mbox{mA}}{\mbox{pulse}} \times 300 \frac{\mbox{pulses}}{\mbox{sec} \times 0.1 \mu \mbox{s}[/math]

Time on for one second is 30 us. and Time off is 0.999999 s The average current 0.999999 * 0 + 0.000001*300*200mA = 60[math]\mu[/math]A.

P = (10 MeV) [math]\times[/math] (60 [math]\mu[/math]A) = 600 Watts.

Properties of Tungsten

  • Melting Point = 3695 K.
  • Heat Capacity = [math](25~ ^{o}C)~ 24.27 ~ J ~ mol^{-1} ~ K^{-1}[/math]

  • Heat loss due to radiation: ( \sigma T^4)

Tungsten Temperature as a function of heat load

  • IAC beamline pressure = [math]10^{-8} [/math] Torr
  • The Tungsten heats up when an MeV energy electron impinges its surface. The imperfect vacuum inside the beam pipe does allow some radiative cooling.
  • Conduction and Radiation

Calculating Radiators Equilibrium Temperature

1.Calculating number of particles per second

We have electron beam of:

Frequency: f=1000Hz

Peak current: I=10~mAmp=0.01 Amp

Pulse width: [math] \Delta t= 50~ns=5 \times10^{-8}[/math] seconds

So, how many electrons we have in each second?

By Q=It, we have

                           [math] N \times e=f \times I \times \Delta t [/math]

Where N is the total electron numbers hits target per second, e is electron charge and f, I and ∆t are given above. Number of particle per second is:

                           [math]  N = \frac {f \times I \times \Delta t}{e} [/math]

2.Calculating Energy deposited per second

If we find the energy deposited by each electron and multiply to the total number of electrons in each second, we will find the total energy per second deposited in radiator.

To find energy deposited by each electron, we need to use formula

                              [math] E_{dep~one}={(\frac{dE}{dx})}_{coll}\times t [/math]

Where is [math] E_{dep~one} [/math] is energy deposited by one electron, [math] {(\frac{dE}{dx})}_{coll} [/math] is mean energy loss (also stopping power) by collision of electron and [math] t [/math] is thickness of the radiator.

Actually, energy loss of a electron comes from two parts: 1) the emission of electromagnetic radiation arising from scattering in the electric field of a nucleus (bremsstrahlung); 2) Collisional energy loss when passing through matter. Bremsstrahlung will not contribute to the temperature, because it is radiation.

Stopping power can be found from nuclear data tables [math] (dE/dx)_{ave} [/math] and thickness is 0.001 times of radiation length. From Particle Data group we got radiation length and average total stopping powers around 15MeV for electrons in these materials from National Institute of Standards and Technology: Tungsten Stopping Power.

Table of Radiation Lengths

Note:These data is from Particle Data group,Link: [1].

Elements Radiation Lengths [math] (g/cm^{2} )[/math]
W 6.76

Table of energy calculations

For the thickness of 0.001 Radiation Length (0.0001RL) of radiators. Note: [math](dE/dx)_{coll}[/math] is from National Institute of Standards and Technology. Link: [[2]])

Elements [math](dE/dx)_{coll} (MeV \; cm^2/g)[/math] [math] t~( g~cm^{-2}[/math]) [math]E_{(dep~one)}[/math] (MeV) [math]E_{dep/s}[/math] (MeV/s) [math]E_{dep/s}[/math] (J/s)
W 1.247 0.00676 0.00842972 [math]2.63*10^{10}[/math] [math]4.21*10^{-3}[/math]

In above table,we took the total numbers of electrons per second and multiply it to Energy deposited by one electron,get total energy deposited per second (which is power).

                                         [math]P_{dep}= E_{dep}/s = ( E_{(dep~by~one)})\times(Number~of~electrons~per~second)[/math]

3.Calculating equilibrium temperature using Stefan–Boltzmann law

If we assume that there is no energy conduction and total energy is just radiated from two surfaces of the radiators which are as big as beam spot,in our case beam spot is 2mm in diameter. According to Stefan–Boltzmann law, this total power radiated will be

                                           [math] P_{rad} =  A \sigma T^{4}  [/math]

Where T is radiating temperature P is the radiating power, A is surface area that beam incident and σ is Stefan–Boltzmann constant or Stefan's constant. To reach equilibrium temperature, Power deposited in and power radiated should be. So

                                           [math]  P_{dep}=P_{rad}  [/math]


                                        [math] T = [  \frac{P_{dep}}{A\sigma} ]^{1/4} = [  \frac{N*E_{(dep~one)}}{A\sigma} ]^{1/4} [/math]

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